Three rods of material X and three rods of material Y are connected as shown in the figure. All the rods are of identical length and cross-sectional area. If the end A is maintained at `60^@C` and the junction E at `10^@C`. Calculate the temperature of the junction B, C and D. The thermal conductivity of X is `0.92cal//sec-cm^@C` and that of Y is `0.46cal//sec-cm-^@C`.

Explanation

Let `k_(x)` and `k_(y)` be the thermal conductivities of X and Y respectively and let `T_(B),T_(C)` and `T_(D)` be the temperatures of junctionsB,CandDrespectively. Given `T_(A) =60^(@)C` and`T_(E)= 10^(@)C`.
In order to solvethis problem, wewill applythe principle that, in the steady state, the rate at which heat enters a junction is equal to the rate at which heat leavesthat junction. In Figure- 4.18 the direction offlow ofheat is given by arrows.
For junction B, we have
`(k_(y)A(T_(A)-T_(B)))/(d) = (k_(x)A(T_(B)-T_(C)))/(d) + (k_(y)A(T_(B)-T_(D)))/(d)`
or `k_(y)(T_(A)-T_(B)) =k_(x)(T_(B)-T_(C))+k_(y)(T_(B)-T_(D))`
or `(T_(A)-T_(B)) =(k_(x))/(k_(y))(T_(B)-T_(C))+k_(y)(T_(B)-T_(D))`
Given `(k_(x))/(k_(y))=(9.2 xx 10^(-2))/(4.6 xx 10^(-2)) = 2`
and `T_(A)=60^(@)C`.
Therefore, `(60 -T_(B)) = 2(T_(B)-T_(C))+(T_(B)-T_(D))`
or `4T_(B)-2T_(C)-T_(D) = 60`
For junction C, we have
`(k_(x)A(T_(B)-T_(C)))/(d) =(k_(x)A(T_(C)-T_(D)))/(d) + (k_(x)A(T_(C)-T_(E)))/(d)`
Solving we get
`-T_(B)+3T_(C)-T_(D)=10 ` `[At T_(E)=10^(@)]` ....(4.26)
For junction D, we have
or `(k_(y)A(T_(B)-T_(D)))/(d)+(k_(x)A(T_(C)-T_(D)))/(d) = (k_(Y)A(T_(D)-T_(E)))/(d)`
or `(T_(B)-T_(D))+(k_(x))/(k_(y))(T_(C)-T_(D))+(T_(D)-T_(E))`
Putting `(k_(x))/(k_(y))=2` and `T_(E)=10^(@)C` in this equation , we have
`(T_(B)-T_(D))+2(T_(C)-T_(D))=(T_(D)-10)`
`T_(B)+2T_(C)-4T_(D)=-10` ...(4.27)
Solving equation (4.25),(4.26) and (4.27), we get
`T_(B)= 30^(@)C`
and `T_(C)=T_(D) = 20^(@)C` .